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\title{\heiti\zihao{2} 复变函数-第7章}
\author{20373963-樊若宸}
\date{\today}

\begin{document}
\maketitle
\textbf{8.}求下列函数的傅里叶变换

(1)$f(t)=\cos t \sin t$

\textbf{解:}\quad
$$
\begin{aligned}
    \int_{-\infty}^{+\infty}f(t)\mathrm{e}^{-i\omega t}\mathrm{d}t&=\int_{-\infty}^{+\infty}\dfrac{1}{2}\sin 2t\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\dfrac{1}{4i}\int_{-\infty}^{+\infty}[\mathrm{e}^{2it}-\mathrm{e}^{-2it}]\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\dfrac{\pi}{2i}\left[\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}\mathrm{e}^{-i(\omega-2)t}-\mathrm{e}^{-i(\omega+2)t}\mathrm{d}t\right]\\
    &=\dfrac{\pi}{2i}\delta(\omega-2)-\dfrac{\pi}{2i}\delta(\omega+2 )
\end{aligned}
$$

(2)$f(t)=\sin^3 t$

\textbf{解:}\quad
$\sin^3 t = \dfrac{1}{4}\left(3\sin t - \sin 3t\right)$,从而
$$
\begin{aligned}
    F[f] = \dfrac{1}{4}\left[\dfrac{3\pi}{i}\delta(\omega - 1)-\dfrac{3\pi}{i}\delta(\omega+1)-\dfrac{\pi}{i}\delta(\omega-3)+\dfrac{\pi}{i}\delta(\omega+3)\right]
\end{aligned}
$$



(3)$f(t)=\delta(t+a)+\delta(t-a)+\delta(t+\dfrac{a}{2})+\delta(t-\dfrac{a}{2}))$

\textbf{解:}\quad
$$
\begin{aligned}
    F[f]&=\mathrm{e}^{i\omega a}+\mathrm{e}^{-i\omega a}+\mathrm{e}^{i\omega a/2}+\mathrm{e}^{-i\omega a/2}\\
    &=2\cos a\omega + 2\cos\dfrac{a\omega}{2}
\end{aligned}
$$

(4)$f(t)=\sin\left(5t+\dfrac{\pi}{2}\right)$

\textbf{解:}\quad
$F[\sin(5t)] = \dfrac{\pi}{i}\left[\delta(\omega - 5)-\delta(\omega+5)\right]$,从而

$$
\begin{aligned}
    F[f]&=\mathrm{e}^{i\pi/10}\dfrac{\pi}{i}\left[\delta(\omega - 5)-\delta(\omega+5)\right]
\end{aligned}
$$



\textbf{9.}求下列函数的傅里叶逆变换

(1)$F(\omega)=\pi[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]$

\textbf{解:}\quad
$$
\begin{aligned}
    F^{-1}[F(\omega)]&=\dfrac{1}{2}\int_{-\infty}^{+\infty}\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\mathrm{e}^{i\omega t}\mathrm{d}\omega\\
    &=\dfrac{\mathrm{e}^{i\omega_0 t}+\mathrm{e}^{-i\omega_0 t}}{2}\\
    &=\cos \omega_0 t
\end{aligned}
$$

(2)$F(\omega)=\dfrac{2A\sin\omega}{\omega}(A\neq 0)$

\textbf{解:}\quad
$$
\begin{aligned}
    F^{-1}[f]&=\dfrac{A}{\pi}\int_{-\infty}^{+\infty}\dfrac{\sin\omega}{\omega}(\cos\omega t + i\sin\omega t)\mathrm{d}\omega\\
    &=\dfrac{2A}{\pi}\int_{0}^{+\infty}\dfrac{\sin\omega}{\omega}\cos\omega t\mathrm{d}\omega\\
    &=\left\{\begin{array}{ll}
        A & |t|< 1\\
        \dfrac{A}{2} & |t|=1\\
        0 & |t|>1
    \end{array}\right.
\end{aligned}
$$

(3)$\sin(\omega t_0)$

\textbf{解:}\quad
$$
\begin{aligned}
    F^{-1}[\sin(\omega t_0)]&=\dfrac{1}{4\pi i}\int_{-\infty}^{+\infty}\mathrm{e}^{i\omega t_0}-\mathrm{e}^{-i\omega t_0}\cdot\mathrm{e}^{i\omega t}\mathrm{d}\omega\\
    &=\dfrac{1}{2i}\left[\delta(t+t_0)-\delta(t-t_0)\right]
\end{aligned}
$$

\end{document}